基于参数特征的部分模板专业化(Partial template specialization based on argument traits)

基于参数特征的部分模板专业化(Partial template specialization based on argument traits)

假设我有以下模板：

template <typename T> union example { T t; constexpr example(const T & t) : t(t) {}; /* We rely on owning class to take care * of destructing the active member */ ~example() {}; };

由于那里有析构函数， example<T>永远不会被轻易破坏（因此不是文字类型）。 我想要像我这样的部分专业化

template <typename T> union example<std::enable_if_t<std::is_trivially_destructible<T>::value, T>> { T t; constexpr example(const T & t) : t(t) {}; };

让example<T>可以轻易破坏，但不幸的是，这给了我（合理的，事后看来）的警告

警告：类模板部分特化包含一个无法推导出的模板参数; 永远不会使用这种部分专业化

那么有什么方法可以得到我想要的东西吗？

Suppose I have the following template:

template <typename T> union example { T t; constexpr example(const T & t) : t(t) {}; /* We rely on owning class to take care * of destructing the active member */ ~example() {}; };

Because of the destructor there, example<T> will never be trivially destructible (and thus not, say, a literal type). I'd like to have a partial specialization like

template <typename T> union example<std::enable_if_t<std::is_trivially_destructible<T>::value, T>> { T t; constexpr example(const T & t) : t(t) {}; };

to let example<T> be trivially destructible when T is, but unfortunately that gives me the (reasonable, in hindsight) warning

warning: class template partial specialization contains a template parameter that can not be deduced; this partial specialization will never be used

So is there any way to get what I want here?

最满意答案

也许使用第二个默认模板参数？

#include <type_traits> #include <iostream> template <typename T, bool = std::is_trivially_destructible<T>::value> union example { T t; constexpr example(const T & t) : t(t) {}; /* We rely on owning class to take care * of destructing the active member */ ~example() { std::cout << "primary template\n"; } }; template<typename T> union example<T, true> { T t; constexpr example(const T & t) : t(t) {}; }; struct nontrivial { ~nontrivial() { std::cout << "woot!\n"; } }; int main() { example<nontrivial> e1{{}}; example<int> e2{{}}; }

Maybe with a second, defaulted template parameter?

#include <type_traits> #include <iostream> template <typename T, bool = std::is_trivially_destructible<T>::value> union example { T t; constexpr example(const T & t) : t(t) {}; /* We rely on owning class to take care * of destructing the active member */ ~example() { std::cout << "primary template\n"; } }; template<typename T> union example<T, true> { T t; constexpr example(const T & t) : t(t) {}; }; struct nontrivial { ~nontrivial() { std::cout << "woot!\n"; } }; int main() { example<nontrivial> e1{{}}; example<int> e2{{}}; }